parametrization - cornell university
TRANSCRIPT
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Parametrization
Dr. Allen Back
Nov. 17, 2014
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
The graph z = F (x , y) can always be parameterized by
Φ(u, v) =< u, v ,F (u, v) > .
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
The graph z = F (x , y) can always be parameterized by
Φ(u, v) =< u, v ,F (u, v) > .
Parameters u and v just different names for x and y resp.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
The graph z = F (x , y) can always be parameterized by
Φ(u, v) =< u, v ,F (u, v) > .
Use this idea if you can’t think of something better.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
The graph z = F (x , y) can always be parameterized by
Φ(u, v) =< u, v ,F (u, v) > .
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
The graph z = F (x , y) can always be parameterized by
Φ(u, v) =< u, v ,F (u, v) > .
Note the curves where u and v are constant are visible in thewireframe.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
A trigonometric parametrization will often be better if you haveto calculate a surface integral.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
A trigonometric parametrization will often be better if you haveto calculate a surface integral.
Φ(u, v) =< 2u cos v , u sin v , 4u2 > .
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
A trigonometric parametrization will often be better if you haveto calculate a surface integral.
Φ(u, v) =< 2u cos v , u sin v , 4u2 > .
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
A trigonometric parametrization will often be better if you haveto calculate a surface integral.
Φ(u, v) =< 2u cos v , u sin v , 4u2 > .
Algebraically, we are rescaling the algebra behind polarcoordinates where
x = r cos θ
y = r sin θ
leads to r2 = x2 + y2.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
A trigonometric parametrization will often be better if you haveto calculate a surface integral.
Φ(u, v) =< 2u cos v , u sin v , 4u2 > .
Here we want x2 + 4y2 to be simple. So
x = 2r cos θ
y = r sin θ
will do better.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Paraboloid z = x2 + 4y 2
A trigonometric parametrization will often be better if you haveto calculate a surface integral.
Φ(u, v) =< 2u cos v , u sin v , 4u2 > .
Here we want x2 + 4y2 to be simple. So
x = 2r cos θ
y = r sin θ
will do better.Plug x and y into z = x2 + 4y2 to get the z-component.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Parabolic Cylinder z = x2
Graph parametrizations are often optimal for paraboliccylinders.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Parabolic Cylinder z = x2
Φ(u, v) =< u, v , u2 >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Parabolic Cylinder z = x2
Φ(u, v) =< u, v , u2 >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Parabolic Cylinder z = x2
Φ(u, v) =< u, v , u2 >
One of the parameters (v) is giving us the “extrusion”direction. The parameter u is just being used to describe thecurve z = x2 in the zx plane.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
The trigonometric trick is often good for elliptic cylinders
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
Φ(u, v) =<√
3·√
2 cos v , u,√
3 sin v >=<√
6 cos v , u,√
3 sin v >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
Φ(u, v) =<√
6 cos v , u,√
3 sin v >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
Φ(u, v) =<√
6 cos v , u,√
3 sin v >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
Φ(u, v) =<√
6 cos v , u,√
3 sin v >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
Φ(u, v) =<√
6 cos v , u,√
3 sin v >
What happened here is we started with the polar coordinateidea
x = r cos θ
z = r sin θ
but noted that the algebra wasn’t right for x2 + 2z2 so shiftedto
x =√
2r cos θ
z = r sin θ
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Elliptic Cylinder x2 + 2z2 = 6
Φ(u, v) =<√
6 cos v , u,√
3 sin v >
x =√
2r cos θ
z = r sin θ
makes the left hand side work out to 2r2 which will be 6 whenr =√
3.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Ellipsoid x2 + 2y 2 + 3z2 = 4
A similar trick occurs for using spherical coordinate ideas inparameterizing ellipsoids.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Ellipsoid x2 + 2y 2 + 3z2 = 4
A similar trick occurs for using spherical coordinate ideas inparameterizing ellipsoids.
Φ(u, v) =< 2 sin u cos v ,√
2 sin u sin v ,
√4
3cos u >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Ellipsoid x2 + 2y 2 + 3z2 = 4
Φ(u, v) =< 2 sin u cos v ,√
2 sin u sin v ,
√4
3cos u >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperbolic Cylinder x2 − z2 = −4
You may have run into the hyperbolic functions
cosh x =ex + e−x
2
sinh x =ex − e−x
2
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperbolic Cylinder x2 − z2 = −4
You may have run into the hyperbolic functions
cosh x =ex + e−x
2
sinh x =ex − e−x
2
Just as cos2 θ + sin2 θ = 1 helps with ellipses, the hyperbolicversion cosh2 θ − sinh2 θ = 1 leads to the nicest hyperbolaparameterizations.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperbolic Cylinder x2 − z2 = −4
Just as cos2 θ + sin2 θ = 1 helps with ellipses, the hyperbolicversion cosh2 θ − sinh2 θ = 1 leads to the nicest hyperbolaparameterizations.
Φ(u, v) =< 2 sinh v , u, 2 cosh v >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperbolic Cylinder x2 − z2 = −4
Φ(u, v) =< 2 sinh v , u, 2 cosh v >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Saddle z = x2 − y 2
The hyperbolic trick also works with saddles
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Saddle z = x2 − y 2
Φ(u, v) =< u cosh v , u sinh v , u2 >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Saddle z = x2 − y 2
Φ(u, v) =< u cosh v , u sinh v , u2 >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperboloid of 1 Sheet x2 + y 2 − z2 = 1
The spherical coordinate idea for ellipsoids with sinφ replacedby cosh u works well here.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperboloid of 1 Sheet x2 + y 2 − z2 = 1
Φ(u, v) =< cosh u cos v , cosh u sin v , sinh u >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperboloid of 1 Sheet x2 + y 2 − z2 = 1
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperboloid of 2-Sheets x2 + y 2 − z2 = −1
Φ(u, v) =< sinh u cos v , sinh u sin v , cosh u >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Hyperboloid of 2-Sheets x2 + y 2 − z2 = −1
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Top Part of Cone z2 = x2 + y 2
So z =√
x2 + y2.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Top Part of Cone z2 = x2 + y 2
So z =√
x2 + y2.The polar coordinate idea leads to
Φ(u, v) =< u cos v , u sin v , u >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Top Part of Cone z2 = x2 + y 2
So z =√
x2 + y2.The polar coordinate idea leads to
Φ(u, v) =< u cos v , u sin v , u >
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Mercator Parametrization of the Sphere
For 0 ≤ v ≤ ∞, 0 ≤ u ≤ 2π
Φ(u, v) = (sech(v) cos u, sech(v) sin u, tanh(v)).
(Note tanh2(v) + sech2(v) = 1)
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Picture of ~Tu, ~Tv for a Lat/Long Param. of the Sphere.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Basic Parametrization Picture
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Parametrization Φ(u, v) = (x(u, v), y(u, v), z(u, v))
Tangents Tu = (xu, yu, zu) Tv = (xv , yv , zv )
Area Element dS = ‖~Tu × ~Tv‖ du dv
Normal ~N = ~Tu × ~Tv
Unit normal n̂ = ±~Tu × ~Tv |‖~Tu × ~Tv‖
(Choosing the ± sign corresponds to an orientation of thesurface.)
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Two Kinds of Surface Integrals
Surface Integral of a scalar function f (x , y , z) :∫∫S
f (x , y , z) dS
Surface Integral of a vector field ~F (x , y , z) :∫∫S~F (x , y , z) · n̂ dS .
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Surface Integral of a scalar function f (x , y , z) calculated by∫∫S
f (x , y , z) dS =
∫∫D
f (Φ(u, v)) ‖~Tu × ~Tv‖ du dv
where D is the domain of the parametrization Φ.Surface Integral of a vector field ~F (x , y , z) calculated by∫∫
S~F (x , y , z) · n̂ dS
= ±∫∫D~F (Φ(u, v)) ·
(~Tu × ~Tv |‖~Tu × ~Tv‖
)‖~Tu × ~Tv‖ du dv
where D is the domain of the parametrization Φ.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
3d Flux Picture
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
The preceding picture can be used to argue that if ~F (x , y , z) isthe velocity vector field, e.g. of a fluid of density ρ(x , y , z),then the surface integral∫∫
Sρ~F · n̂ dS
(with associated Riemann Sum∑ρ(x∗i , y
∗j , z∗k )~F (x∗i , y
∗j , z∗k ) · n̂(x∗i , y
∗j , z∗k ) ∆Sijk)
represents the rate at which material (e.g. grams per second)crosses the surface.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
From this point of view the orientation of a surface simple tellsus which side is accumulatiing mass, in the case where thevalue of the integral is positive.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
2d Flux Picture
There’s an analagous 2d Riemann sum and interp of∫C~F · n̂ ds.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Problem: Calculate ∫∫S~F (x , y , z) · n̂ dS
for the vector field ~F (x , y , z) = (x , y , z) and S the part of theparaboloid z = 1− x2 − y2 above the xy -plane. Choose thepositive orientation of the paraboloid to be the one with normalpointing downward.
Parametrization
V1
SurfaceParametriza-tion
SurfaceIntegrals
Surface Integrals
Problem: Calculate the surface area of the above paraboloid.